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Distance between 2 atoms in a crystal

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Given a crystal (cube) with regularly arranged atoms (Total 27 atoms)
    The mass of each atom is 3.5*10^-25. The density of the crystal is 9.2*10^3
    What is the shortest distance between the centres of two adjacent atoms?

    2. Relevant equations

    density = mass / volume

    3. The attempt at a solution

    Volume= (27*3.5*10^-25)/(9.2*10^3)=1.03*10^-27
    Length(L)=1*10-9
    Distance between two atoms is L/2=5.04*10^-10
    Correct answer is 3.4*10^-9
     
  2. jcsd
  3. Jun 9, 2013 #2
    do you thinkthis is the correct answer?
     
  4. Jun 9, 2013 #3
    I think your "correct answer" is off by 10
     
  5. Jun 9, 2013 #4
    L is correct. Should you really be dividing by 2?
     
  6. Jun 9, 2013 #5
    Sorry the correct answer should be 3.4*10^-10
    Should I divide by 3 then ? Why ?
     
  7. Jun 9, 2013 #6
    What you probably did was to assume the volume of 27 atoms had an atom at every corner and one between. But if you put this volume next to another, the atoms on the edges would be to close. In fact, the distance should be 3.
     
  8. Jun 11, 2013 #7
    So the atoms on the edges would be closer than others and that's the required "shortest distance"?
    I just can't really imagine what it looks like
     
  9. Jun 11, 2013 #8
    Maybe I wasn't clear. Put the atoms in a uniform rectangular three dimensional grid. My attachment shows the top view. Notice the volume is (3d)^3
     

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  10. Jun 11, 2013 #9

    haruspex

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    The difficulty is, what does volume (and therefore density) mean when applied to a crystal consisting of only 27 atoms? It does not say that the density given is for arbitrarily large chunks of this material - it is for specifically this crystal of 27 atoms. So I tend to agree with Nemo's answer.
     
  11. Jun 11, 2013 #10
    Good point, I guess whoever stated the "correct answer" thought it was for a large chunk.
     
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